Solve for $x$ and $y$ by deriving an expression for $x$ from the second equation, and substituting it back into the first equation. $\begin{align*}3x+7y &= 5 \\ -6x-2y &= -2\end{align*}$
Explanation: Begin by moving the $y$ -term in the second equation to the right side of the equation. $-6x = 2y-2$ Divide both sides by $-6$ to isolate $x$ $x = {-\dfrac{1}{3}y + \dfrac{1}{3}}$ Substitute this expression for $x$ in the first equation. $3({-\dfrac{1}{3}y + \dfrac{1}{3}}) + 7y = 5$ $-y + 1 + 7y = 5$ Simplify by combining terms, then solve for $y$ $6y + 1 = 5$ $6y = 4$ $y = \dfrac{2}{3}$ Substitute $\dfrac{2}{3}$ for $y$ in the top equation. $3x+7( \dfrac{2}{3}) = 5$ $3x+\dfrac{14}{3} = 5$ $3x = \dfrac{1}{3}$ $x = \dfrac{1}{9}$ The solution is $\enspace x = \dfrac{1}{9}, \enspace y = \dfrac{2}{3}$.